# Commutative Banach algebras and maximum ideal space

Let $$A, B$$ be commutative unital Banach algebras and let $$\varphi: A \rightarrow B$$ be a continuous unital map such that $$\overline{\varphi(A)} = B$$

Let $$\varphi^{*}: \text{Max}(B) \rightarrow \text{Max}(A)$$ $$\varphi^{*}(m) = m(\varphi)$$ be the map from the space of maximal ideals of $$B$$ to the space of maximal ideals of $$A$$ induced by $$\varphi$$.

How to prove that $$\varphi^{*}$$ is a topologically injective map?

(Recall that an operator $$T: X \rightarrow Y$$ is called topologically injective if $$T: X \rightarrow \text{im}(T)$$ is a homeomorphism)

My progress on the problem is the following: first of all, the space of maximal ideals of a commutative Banach algebra $$A$$ can be identified with the space of continuous functionals of the form $$m: A \rightarrow \mathbb{C}$$. Clearly the map above is continuous, since the pointwise convergence of a net $$(n_{i})$$ in $$\text{Max}(B)$$ implies the convergence of the net $$m_{i} \circ \varphi)$$

(the space of continuous linear functional is endowed with the weak* topology)

If we assume for the moment that the map is bijective, then the fact that a continuous bijective map between compact Hausdorff spaces is a homeomorphism, yields the result.

(here the space of maximal ideals is compact in weak* topology since the algebra is unital)

The suggested proposition looks like a relaxation of the aformentioned reasoning above though i cannot figure out an easy way to modify it to make it work. Are there any hints?

• You need to use the fact that all characters in the spectrum are of norm one to relate the topology of $\mathrm{im}(\varphi^\ast)$ with the topology of $\mathrm{Max}(B)$. – Adrián González-Pérez Dec 26 '18 at 13:45

Indeed, they are bounded since their kernel is a maximal ideal and therefore closed. The fact that they are contractive follows from spectral theory. If $$a - \lambda 1$$ is invertible so is $$\chi(a) - \lambda$$. The counter-reciprocal gives that when $$\lambda$$ is in the image of $$a$$ under $$\chi$$ then $$a - \lambda 1$$ is not invertible and so $$|\chi(a)| \leq \sup \{ |\lambda| : \lambda \in \mathrm{sp}(a) \} \leq \| a \|$$
Using the fact that $$\varphi[A]$$ is dense in $$B$$ you have that if two continuous functional $$\varphi_1$$ and $$\varphi_2$$ agree on $$\varphi[A]$$, then they are equal. This gives the injectivity.
For topological injectivity you need to see that if $$\varphi^\ast(\chi_n) = \chi_n \circ \varphi \in \mathrm{im}(\varphi^*)\subset \mathrm{Max}(A)$$ converge to $$\chi \circ \varphi$$ then $$\chi_n \to \chi$$ in $$\mathrm{Max}(B)$$. Let $$b \in B$$, we can find $$a_\epsilon$$ with $$\|\varphi(a_\epsilon) - b\| < \epsilon$$ and $$\begin{eqnarray*} |\chi_n(b) - \chi(b) | & \leq &\| \chi_n(b) - \chi_n(\varphi(a_\epsilon))\| + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + | \chi(\varphi(a_\epsilon)) - \chi(b)|\\ & \leq & \epsilon + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + \epsilon. \end{eqnarray*}$$ But that implies that the limit of $$|\chi_n(b) - \chi(b)|$$ is smaller or equal than $$\epsilon$$ for every $$\epsilon$$ and therefore $$0$$.